Question

At a local fair, the admission fees are $1.50 for children, $3 for college students, and $5 for adults. On Saturday, 1500 tickets were sold and brought in $4950. The number of college student tickets sold were 100 less than the number of children's tickets sold. How many children's tickets were sold? ticket

Answer 1

Answer:

500

Step-by-step explanation:

Let's say x represents the number of children's tickets sold, y represents the number of college tickets sold, and z represents the number of adult tickets sold.

For each child ticket sold, we add x to the total. Similarly, for college tickets, we add y and for adult tickets, we add z.

The total revenue can therefore be represented as

(child ticket revenue) + (college ticket revenue) + (adult ticket revenue) =

(price * quantity of children's tickets) + (price * quantity of college tickets) + (price * quantity of adult tickets) =

(1.5 * x) + (3 * y) + (5 * z) = 4950

The total amount of tickets is x+y+z = 1500

The number of college student tickets (y) is 100 less than the number of children's tickets (x), so y = x - 100.

Put these three equations together to get

(1.5 * x) + (3 * y) + (5 * z) = 4950

x+y+z = 1500

y = x-100

Substitute x-100 for y in the second equation

x+(x-100) + z = 1500

add 100 to both sides to isolate the variables

2x + z = 1600

subtract 2x from both sides to isolate the z

1600 - 2x = z

Substitute 1600-2x for z and x-100 for y in the first equation

(1.5 * x) + (3 * y) + (5 * z) = 4950

1.5x + 3*(x-100) + 5 * (1600-2x) = 4950

= 1.5x + 3x - 300 + 8000 - 10x

= 7700 - 5.5x

subtract 7700 from both sides to isolate the x and its coefficient

-2750 = -5.5x

divide both sides by -5.5 to isolate the x

x = 500

Answer 2

In this exercise we have to use the knowledge of finance to calculate the amount of tickets sold to children based on the profit made, in this way we can say that this value corresponds to:

500  children's tickets.

Let's say x represents the number of children's tickets sold, y represents the number of college tickets sold, and z represents the number of adult tickets sold.

For each child ticket sold, we add x to the total. Similarly, for college tickets, we add y and for adult tickets, we add z.

The total revenue can therefore be represented as:

$(child \ ticket\ revenue) + (college\ ticket\ revenue) + (adult\ ticket \ revenue) = \\=(price * children's \ tickets) + (price * college \ tickets) + (price * adults\ tickets)$

Replacing the values ​​already known in the text in the formula given above, we have:

$(1.5 * x) + (3 * y) + (5 * z) = 4950\\\\\left \{ {{x+y+z = 1500} \atop {y = x-100} \right.$

Substitute x-100 for y in the second equation

$x+(x-100) + z = 1500\\2x + z = 1600\\1600 - 2x = z\\(1.5 * x) + (3 * y) + (5 * z) = 4950\\1.5x + 3*(x-100) + 5 * (1600-2x) = 4950\\= 1.5x + 3x - 300 + 8000 - 10x\\= 7700 - 5.5x$

subtract 7700 from both sides to isolate the x and its coefficient

$-2750 = -5.5x\\x = 500$

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